SRMJEE Mock Test - 7 (Engineering) - JEE MCQ

Field at the centre of coil B' =
Or since R is 2R;
Magnetic field at a distance 6R from the centre of the coil
B =
B =
Or B =
Or ratio =

Velocity of the electron orbiting around the nucleus
v ∝ 1/n _____ 1
and
Radius of the orbit
r ∝ n² ____ 2
Magnetic moment (μ) = (mvr)/2 (m is mass of the electron which is constant) ........... (3)
Therefore,
From 1, 2 and 3 we have
μ ∝ n

For series combination,

= 12 k V

Given: magnetic moment, M = 1.0 Am², separation d = 2 m.

Since axes are perpendicular, so mid-point lies on the axial line of one magnet and on the equatorial line of other magnet.

and 

B₁⊥ B₂

∴ Resultant magnetic field

= 

 
=  Constant

Given:

• Pressure of gas: P = 4.5 × 10³ Pa
• Heat supplied to the gas: ΔQ = 800 kJ = 8 × 10⁵ J
• Initial Volume: V₁ = 0.5 m³
• Final Volume: V₂ = 2.0 m³

Now, work done by the gas in expanding:
ΔW = P(V₂ − V₁)
ΔW = 4.5 × 10⁵ (2.0 − 0.5)
ΔW = 4.5 × 10⁵ × 1.5
ΔW = 6.75 × 10⁵ J

Now, from the first law of thermodynamics:

ΔU = ΔQ − ΔW
where ΔU is the change in internal energy of the gas.

ΔU = 8 × 10⁵ − 6.75 × 10⁵ J
ΔU = 1.25 × 10⁵ J

Thus, the change in internal energy of the gas is 1.25 × 10⁵ J.

The conversion of RCOOH to RCH₂OH can be obtained by lithium aluminium hydride.
NaBH₄ does not reduce carboxylic acids.
Red P/HI would reduce it to the hydrocarbon stage.
Zn-Hg/conc. HCl: Clemmensen reduction is done for conversion of aldehydes to hydrocarbons

Stability order : II > I > III
Note: In III, carbocation is localised due to S.I.R. effect
S.I.R, is stearic inhibition in resonance.

The chemical name of SO₂Cl₂ is sulphuryl chloride.

Structure and Hybridization:
The total number of sigma bonds formed by sulphur in SO₂Cl₂ is four, and there is no lone pair of electrons.
Hence, the hybridization of S in SO₂Cl₂ is sp³.

Oxidation State of S in SO₂Cl₂:
Let the oxidation state of S be x:
x + 2(-2) + 2(-1) = 0
x - 4 - 2 = 0
x = +6
Thus, the oxidation state of S in SO₂Cl₂ is +6.

The chemical name of SOCl₂ is thionyl chloride.

a

Structure and Hybridization:
The total number of sigma bonds formed by sulphur in SOCl₂ is three, and there is one lone pair of electrons.
Hence, the hybridization of S in SOCl₂ is sp³.
Oxidation State of S in SOCl₂:
Let the oxidation state of S be x:
x - 2 + 2(-1) = 0
x - 2 - 2 = 0
x = +4
Thus, the oxidation state of S in SOCl₂ is +4.

Given: sin⁻¹(1 − x) − 2sin⁻¹x = π/2
Let x = sin y
∴ sin⁻¹(1 − siny) − 2y = π/2
⇒ sin⁻¹(1 − siny) = π/2 + 2y
⇒ 1 − siny = sin
⇒ 1 − siny = cos2y
⇒ 1 − siny = 1 − 2sin²y as cos2y = 1 − 2sin²y
⇒ 2sin²y − siny = 0
⇒ 2x² − x = 0
⇒ x(2x − 1) = 0
⇒ x = 0, 2x − 1 = 0
⇒ x = 0, 2x = 1
⇒ x = 0, x = 1/2; but x = 1/2 does not satisfy the given equation.
∴ x = 0 is the solution of the given equation.

Given,

f(x) = x³/8 - sin(πx) + 4

Now, x³/8 is a polynomial which is continuous in R and sin(πx) is also continuous everywhere in R.

Therefore, function f(x) is continuous in the interval x ∈ [-4, 4].

Now,

f(-4) = (-4)³/8 - sin(4π) + 4
= -4

f(4) = 4³/8 - sin(4π) + 4
= 12

⇒ f(x) can take value in the range [-4, 12] as f(x) is a continuous function.

∴ f(x) cannot take the value 18.

Using 2sinAcosA = sin2 A, we get

Using, cos(90°−θ) = sinθ, sin(180° − θ) = sinθ, we get

Now using
 we get

Again, using cos(90°−θ) = sinθ,  we get

Again, using  we get

We know that:

|Z|² = Z * Z̄

Now, let Z = x + iy, and the given inequality is equivalent to:

(2Z − 3i)(2Z̄ + 3i) < (3Z − 2i)(3Z̄ + 2i)

Expanding both sides:

4ZZ̄ + 9 < 9ZZ̄ + 4

Rearranging:

4|Z|² + 9 < 9|Z|² + 4

Simplifying:

5 < 5|Z|²

Dividing both sides by 5:

1 < |Z|²

Taking the square root:

√(x² + y²) > 1

Squaring both sides:

x² + y² > 1

This represents the exterior of the unit circle with its center at (0,0) or Z = 0.

Thus, the correct answer is option B: The exterior of the unit circle with its center at Z = 0.

Given, sin(2x + π/18) · cos(2x − π/9) = -1/4

Multiplying LHS and RHS by 2.

Now, general solution of  sinx = −1 is (4n−1)π/2, n∈Z

Similarly we can find other values of 'x' for different values of  'n' 
∴ In [0,2π] there are four solutions.

Function f(x) is continuos at x = a if

Now, 

Since, the value of sin 1/x is between [−1,1]

= (0)² × (value between −1 and 1)

= 0

f(x)  is continuous at x = 0

if 

⇒ f(0) = 0

Number of persons at least reads two newspapers out of 50 are 12 + 8 + 5 + 2 = 27
So, number of persons reading at least  two newspapers out of 1000 are = 27/50 × 10000 = 5400
Hence, The correct answer is option D.

As we know that, 

Distance between the directrix = 2ae

So, now we need to find the values of a and e.
For that, convert the given equation into standard form or basic definition form.

The given equation is:

(4x − 8)² + 16y² = (x + 3√ − y + 10)²

Rewriting it:

(x − 2)² + y² = [(x + 3√ − y + 10)²] / 16

(x − 2)² + y² = (1 / 2²) * (x + 3√ − y + 10)²

which is of the form PS² = e² PF²,
where P is a point on the ellipse, S is the focus, and F is the foot of the perpendicular from P to the directrix.

Focus = (2, 0)
e = 1/2
Directrix: x + 3√ − y + 10 = 0
Perpendicular distance from focus to directrix:

|2 + 0 + 10| / 2 = 2a − a / 2

12 / 2 = 2a - a / 2

3a / 2 = 6

a = 4

Thus,
Distance between the directrices = 2ae = 2 × 4 × 1/2 = 16

As we know that we can derange r objects out of n objects in ways.
Therefore, the required number of ways

Given that :

∴ Range of f(x) is: 

= 

According to the given information, girls are sitting from left to right as given below.
Neha Shalini Rekha Neetu Pooja
So, Rekha sits in the middle.

The movements of Rohan are shown in the figure below:

Clearly, AD = BC = 2 km
So, required distance = AE
= (DE - AD)
= (3 - 2) km = 1 km

From the above diagram AF = 1 km.

The given word - MEDIATION
First letter of the word - M
Second letter of the word - E
Fifth letter of the word - A
Ninth letter of the word - N
New words that can be made with these four letters - MEAN, NAME
As more than one such words can be made,  'Z' is the answer.
Thus, option D is the correct answer.

It is given that;

Selling price (SP) = 540 Rupees

Loss = 10%

We know that,

In the case of 10% profit gain,

SP =  600 × 110 / 100 = 660 Rupees

'Keep the wolf from the door' means 'have enough money to avert hunger or starvation' 
'Starvation' is a noun which means 'suffering or death caused by lack of food.'
For example: I work part-time to pay the mortgage and keep the wolf from the door
Hence, this is the correct option.